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- Question 9, Exercise 1.2
- e the following formulas: \[\text{Re}(z^{-2}) = \frac{(\text{Re}(z))^2 - (\text{Im}(z))^2}{|z|^4},\] \[\text{Im}(z^{-2}) = \frac{-2 \text{Re}(z) \text{Im}(z)}{|z|^4}. \] First, n... bove formulas \begin{align} Re((3 - 2i)^{-2}) &= \frac{(3)^2 - (-2)^2}{(\sqrt{13})^4} \\ &= \frac{9 - 4}{169} = \frac{5}{169}. \end{align} \begin{align} Im((3
- Question 7, Exercise 1.4
- frac{\pi}{3}$ ** Solution. ** \begin{align*} &-\frac{\pi}{3} \leq \arg (z-4) \leq \frac{\pi}{3}\\ \implies & -\frac{\pi}{3} \leq \arg(x+iy-4) \leq \frac{\pi}{3} \\ \implies & -\frac{\pi}{3} \leq \arg(x-4+iy) \leq \frac{
- Question 1, Exercise 1.4
- ign} and \begin{align} \alpha & = \tan^{-1}\left|\frac{y}{x}\right| = \tan^{-1}\left|\frac{2\sqrt{3}}{2}\right|\\ & = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}. \end{align} Since the complex number \(... e argument \( \theta \) is: \[ \theta = \alpha = \frac{\pi}{3}. \] Hence \[ 2 + i 2 \sqrt{3} = 4 \left(
- Question 6(i-ix), Exercise 1.4
- 10^\circ + i \sin 210^\circ\right) \\ =& 5\left(-\frac{\sqrt{3}}{2} - \frac{1}{2}i\right) \\ =& -\frac{5\sqrt{3}}{2} - \frac{5}{2}i \end{align*} =====Question 6(iii)===== Write a given complex number in the
- Question 10, Exercise 1.2
- +2 i$ and $z_{2}=1-3 i$ verify: $\overline{\left(\frac{z_{1}}{z_{2}}\right)}=\frac{\overline{z_{1}}}{\overline{z_{2}}}$. **Solution.** Given \[z_1 = -3 + 2i, \quad z_2 = 1 - 3i\] Then \begin{align} \frac{z_1}{z_2} &= \frac{-3 + 2i}{1 - 3i}\\ &= \frac{(-3 + 2i)(1 + 3i)}{(1 - 3i)(1 + 3i)} \\ &= \frac{-3 + 2i
- Question 1, Review Exercise
- 4 i$ then $\mathrm{z}^{-1}$ is * (a) $\left(\frac{1}{3}, \frac{1}{4}\right)$ * (b) $\left(-\frac{1}{3},-\frac{1}{4}\right)$ * %%(c)%% $\left(\frac{3}{25}, \frac{-4}{25}\right)$ * (d) $\left(\
- Question 6(x-xvii), Exercise 1.4
- }{4}}$ ** Solution. ** \begin{align*} & 2 e^{i \frac{\pi}{4}} \\ =& 2\left(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}\right) \\ =& 2\left(\frac{1}{\sqrt{2}} + i \cdot \frac{1}{\sqrt{2}}\right) \\ =& \sqrt{2} + \sq
- Question 2, Exercise 1.4
- _1=\cos \dfrac{\pi}{6}+i \sin \dfrac{\pi}{6}=e^{i\frac{\pi}{6}}$ and $z_2=\cos \dfrac{\pi}{3}+i \sin \dfrac{\pi}{3}=e^{i\frac{\pi}{3}}$. Then \begin{align} z_1 z_2 & = e^{i\frac{\pi}{6}} \cdot e^{i\frac{\pi}{3}} \\ & = e^{i\left(\frac{\pi}{6}+\frac{\pi}{3}\right)} \\ & = e^{i\frac
- Question 1, Exercise 1.1
- stion 1(iii)==== Evaluate ${{\left( -1 \right)}^{\frac{-13}{2}}}$. **Solution.** \begin{align}{{\left( -1 \right)}^{\frac{-23}{2}}}&={{\left( \sqrt{-1} \right)}^{-23}}\\ &... ====Question 1(iv)==== Evaluate $\dfrac{2}{(-1)^{\frac{3}{2}}}$. GOOD **Solution.** \begin{align}{{\left( -1 \right)}^{\frac{15}{2}}}&={{\left[ {{\left( -1 \right)}^{\frac{1}
- Question 8, Exercise 1.4
- e^{i\theta} \\ &=0.004 e^{i\dfrac{\pi}{4}} \\ &=\frac{4}{1000} \left(\cos\left(\dfrac{\pi}{4}\right) +i \sin\left(\dfrac{\pi}{4}\right)\right) \\ &=\frac{1}{250} \left(\dfrac{1}{\sqrt{2}} +i \dfrac{1}{\sqrt{2}}\right) \\ &=\frac{1}{250\sqrt{2}}(1 +i) \\ &=\frac{\sqrt{2}}{500}(1 +i). \end{align} Hence mean position of particle is $
- Question 3, Exercise 1.4
- {2}+b^{2}$ (ii) $\sum_{r=1}^{n} \tan ^{-1}\left(\frac{y_{r}}{x_{r}}\right)=\tan ^{-1}\left(\frac{b}{a}\right)+2 k \pi, k \in \mathbb{Z}$ ** Solution. ** Le... {Z}, $$ implies $$\sum_{r=1}^{n} \tan ^{-1}\left(\frac{y_{r}}{x_{r}}\right)=\tan ^{-1}\left(\frac{b}{a}\right)+2 k \pi, k \in \mathbb{Z}. \,\, -- (5)$$ $(4)$
- Question 3, Exercise 1.3
- n 3(ii)==== Solve the quadratic equation: $z^{2}-\frac{1}{2} z+17=0$. **Solution.** Given $$ z^{2} - \frac{1}{2}z + 17 = 0 $$ Using the quadratic formula: $
- Question 8, Exercise 1.2
- as required. GOOD ====Question 8(iv)==== Write $\frac{1}{2} \operatorname{Re}(i \bar{z})=4$ in terms of
- Question 1, Exercise 1.3
- \left(z + \dfrac{3}{2}\right)(2 z^2 - 10) \\ = & \frac{1}{2}(2z + 3)\cdot 2(z^2-5) \\ = & (2z + 3)(z^2-(
- Question 2, Exercise 1.3
- {align} Multiply through by $-2$ to eliminate the fraction: \begin{align} z^2 + 10z - 4 &= 0 \\ z^2 + 1